Question

The chemotherapy drug cisplatin (Pt(NH₃)₂Cl₂) is made by the following reaction:

(NH₄)₂PtCl₄ (s) + NH₃ (aq) $\rightarrow$ NH₄Cl (aq.) + Pt(NH₃)₂Cl₂ (s)

The reaction is not balanced. If you start with 20.0 g of (NH₄)₂PtCl₄ and 2.00 g of NH_3, which reagent is in excess and which reagent is the limiting reagent. Please show plan and set up.

Answer 1

First we have to balance the equation, then we can calculate the moles of reagents and compare in order to know who is in excess:

(NH4)2PtCl4 + 2NH3 -----> 2NH4Cl + Pt(NH3)2Cl2

mol (NH4)2PtCl4 = 20g/371g/mol= 0.054 mol

mol NH3= 2g/17g/mol= 0.118 mol

1 mol (NH4)2PtCl4 -------------- 2 mol NH3

0.054 mol (NH4)2PtCl4 ---------------- x= 0.108 mol NH3 ----> this means that we need 0.108 mol of NH3 in order to completely react with all the (NH4)2PtCl4, but we have 0.118 mol so, the NH3 is in excess, the (NH4)2PtCl4 is the limiting reagent.